Problem: What is the extraneous solution to these equations? $\dfrac{x^2 + 2}{x - 1} = \dfrac{-6x + 9}{x - 1}$
Answer: Multiply both sides by $x - 1$ $ \dfrac{x^2 + 2}{x - 1} (x - 1) = \dfrac{-6x + 9}{x - 1} (x - 1)$ $ x^2 + 2 = -6x + 9$ Subtract $-6x + 9$ from both sides: $ x^2 + 2 - (-6x + 9) = -6x + 9 - (-6x + 9)$ $ x^2 + 2 + 6x - 9 = 0$ $ x^2 - 7 + 6x = 0$ Factor the expression: $ (x + 7)(x - 1) = 0$ Therefore $x = -7$ or $x = 1$ At $x = 1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 1$, it is an extraneous solution.